Synchronous Machines Construction 1

Synchronous Machines Construction 1. Stator - Armature (Same as the stator of an Induction Machine) 2. Rotor - Field Winding ( DC Field) Generator I As a generator, this machine is widely used. I This is the machine which is used in all conventional power plants. I A synchronous generator is called as an Alternator. Motor I Since it runs only at synchronous speed, it is not widely used as a motor but for certain applications like electric clocks. I However, it is used to either produce or absorb reactive power. Construction 1. Stator I It is a distributed winding. I The windings of each phase are distributed over several slots. I Three phase windings are wound 120◦ (electrical angle) apart in space 2. Rotor I It is excited by a DC current. There are two types. 2.1 Cylindrical Rotor - used where the turbine speed is high 2.2 Salient Pole Rotor - used where the turbine speed is low Synchronous machine is a doubly excited machine. We can have field winding placed in stator and armature winding in rotor. This type is called an inverted synchronous machine. Synchrnous Generator B 2π θe 2π θm e !t B 4π(θe ) 2π(θm) e !t In general P θ = θ e 2 m Source: A. E. Fitzgerald, et.al., \ Electric Machinery" v !t In a 2-pole machine, one revolution makes one cycle of alternating voltage. In a P-pole machine, 1 P 1 revolution/min × cycle/sec 60 2 N P N rpm × cycle/sec 60 2 PN f = Hz 120 The flux linkage with stator coil a is λa = Nφf cos !t Where N is the number of turns in each pahse and φf is the flux per pole due to excitation current If . By Faraday's law, the voltage induced in phase a is dλ e = − a = !Nφ sin !t a dt f The rms value in voltage is Ea = 4:44fNφf φf E I The field winding is excited by DC. I When the rotor rotates, the field will revolve in the air gap. I The revolving field induces an emf in the armature winding placed in the stator. I The speed of the field (N) is determined by the frequency of induced voltage. PN f = Hz 120 I Rotor and field due to field winding rotate at the same speed called the synchrnous speed. I When the stator terminals are connected to a load, a three phase armature current flows out of the machine. I When a three phase balanced current flows in the three phase armature winding, a revolving mmf is produced whose speed is 120f N = rpm P I The field due to armature current also rotates at the synchronous speed in the air gap. I Finally the net air gap field is the resultant of field due to If and stator current Ia. Fr = Ff + Fa φr = φf + φa I The resultant and component fluxes rotate in the air gap at the same speed. φf E φr θ E If φa Figure: OCC Ia Figure: Field diagram Equivalent Circuit Since the operation is balanced, the per phase equivalent circuit of a synchronous machine is Xar Xal Ra + + E Vt − − where Xar is the reactance of armature reaction. Xal is the leakage reactance. Ra is the armature resistance. Xs Ra Ia + + E Vt − − The synchronous reactance Xs = Xar + Xal The synchronous impedance Zs = Ra + |Xs E = Vt + Ia(Ra + |Xs ) Determination of Xs Open Circuit Test : R + Field V E If − N V + f − Y B Vary If and measure E. Repeat till E reaches its rated value. Short Circuit Test : Ia R A Field If V + f − Y B Vary If and measure Ia. Repeat till Ia reaches its rated value. The open circuit characteristics is a non linear curve. However, the short circuit characteristics is a straight line because the flux inside the machine is low (no saturation). φf Xs Ra Ia φr + E E − θ φa Figure: SC Test Ia Figure: Field diagram E Ia Eda d OCC V t c SCC I b a a If I Unsaturated synchronous impedance, Eda Zs(unsat) = Ia If Ra is known, Xs(unsat) can be found. q 2 2 Xs(unsat) = (Zs(unsat)) − Ra I Saturated synchronous impedance, Vt Zs(sat) = == Ra + |Xs(sat) Ia If Ra is known, Xs(sat) can be found. q 2 2 Xs(sat) = (Zs(sat)) − Ra Saturated synchronous reactance Xs is normally used in calculations. Alternator - Phasor Diagrams E IaXs δ φ Vt IaRa Ia Figure: lagging power factor load E IaXs δ Ia Vt IaRa Figure: unity power factor load E I X Ia a s δ φ IaRa Vt Figure: leading power factor load In Alternators I E leads Vt by δ. δ is positive. I This δ is also the angle between the main field φf and the resultant field φr . I φf leads φr by δ. Let us neglect Ra. φf φa φr δ E IaXs δ φ Vt Ia Figure: Alternator supplies a lagging power factor load - Field diagram Synchronization of Alternators R RGrid L1 Field Y YGrid If L2 V + f − L3 B BGrid The alternator and the grid must have the same 1. Voltage 2. Frequency 3. Phase Sequence 4. Phase I When all the above conditions are satisfied, the three lamps will be dark. It is time to close the switch. I Once it is synchronized, the terminal voltage and frequency of the alternator can not be changed. I By changing the prime mover speed, Real Power (P) can be controlled. I By changing the excitation, Reactive Power (Q) and hence power factor can be controlled. Power Characteristic It always works with an infinite bus. Xs Ra Ia + + ◦ E δ Vt 0 − − Figure: Per phase equivalent circuit To find the power delivered by the alternator, ∗ S = Vt Ia ◦ ∗ ◦ (E δ − Vt 0 ) S = Vt 0 jZs j θ EV θ − δ V 2 θ S = t − t jZs j jZs j By taking Real and Imaginary parts, EV cos(θ − δ) V 2 cos(θ) P = t − t jZs j jZs j EV sin(θ − δ) V 2 sin(θ) Q = t − t jZs j jZs j ◦ If we neglect Ra, jZs j = jXs j and θ = 90 . EV sin δ P = t Xs EV cos δ V 2 Q = t − t Xs Xs Three phase real and Reactive power EVt sin δ P = 3 = Pmax sin δ Xs EV cos δ V 2 Q = 3 t − t Xs Xs P Pmax δ Figure: Power angle curve V - Curves: Xs Ia + + ◦ E δ Vt 0 − − Figure: Per phase equivalent circuit E IaXs δ φ Vt Ia Figure: Supplying a lagging power factor load Let us keep the real power delivered constant and change the excitation. EVt sin δ P = = Vt Ia cos φ Xs where φ is the angle difference between Vt and Ia. if If is varied, Ef will vary (linearly if magnetic linearity is assumed). Since Vt is fixed, E sin δ = const Ia cos φ = const E2 E E1 Ia2Xs IaXs Ia1Xs δ Ia2 Ia cos φ sin E Ia Vt Ia1 Figure: Different excitation for a constant P E1 sin δ1 = E sin δ = E2 sin δ2 Ia1 cos φ1 = Ia cos φ = Ia2 cos φ2 1. Over Excitation : E cos δ > Vt Alternator supplies lagging current and supplies reactive power. 2. Normal Excitation : E cos δ = Vt Alternator supplies unity power factor current and supplies zero reactive power. 3. Under Excitation : E cos δ < Vt Alternator supplies leading current and draws reactive power. If we plot Ia Vs If , Ia P2 P1 P2 > P1 Under Excitation Over Excitation leading pf lagging pf If Figure: V-Curves of an Alternator Example 1 : (Example 8.13 : Kothari) A 3φ, 10 kVA, 400 V, 4-pole, 50 Hz star connected synchronous machine has synchronous reactance of 16 Ω and negligible resistance. The machine is operating as generator on 400 V bus-bars (assumed infinite). 1. Determine the excitation emf (phase) and torque angle when the machine is delivering rated kVA at 0.8 pf lagging. 2. While supplying the same real power as in part (1), the machine excitation is raised by 20 %. Find the stator current, power factor, torque angle and reactive power delivered. 3. With the field current held constant as in part (1), the power (real) load is increased till the steady state power limit is reached. Calculate the maximum power and kVAR delivered and also the stator current and power factor. Draw the phasor diagram under these conditions. 1. P = 10 × 0:8 = 8kW (three phase) 8000 Ia = p = 14:43 A 3 × 400 × 0:8 400 E δ = p 0◦ + 14:43 −36:87◦ × 16 90◦ 3 E = 413:3 V (phase) δ = 26:5◦ 2. E1 = 1:2 × E = 496 V E sin δ = E1 sin δ1 ◦ δ1 = 21:9 The stator current ◦ E1 δ1 − Vt 0 Ia1 = |Xs ◦ Ia1 = 18:4 −50:9 A pf = 0:63 lagging p Q = 3Vt Ia sin φ = 9:89 kVAR (3-phase) Q can also be found by the following formula. EV cos δ V 2 Q = 3 t − t Xs Xs Q is positive which means the machine is supplying it. 3. E = 413:3 V When the real power reaches its maximum value, δ = 90◦. EVt 413:3 × 231 Pmax = = = 5:96 kW/phase Xs 16 Pmax = 17:38 kW (3-phase) 413:3 90◦ − 231 0◦ I = = 29:56 29:2◦ A a 16 90◦ pf = 0:873 leading p Q = 3×400×29:56×sin(−29:2◦) = −9:99 kVAR (3-phase) Since Q is negative, the machine is drawing it.

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